How to show that following limit is zero?

Question

In one of the research article it is written that the following limit is equal to zero $$\lim_{x \to 0 }\frac{d}{2^{b+c/x}-1}\left[a2^{b+c/x}-a-a\frac{c\ln{(2)}2^{b+c/x}}{2x}-\frac{c\ln{(2)}}{2x^2}\frac{2^{b+c/x}}{\sqrt{2^{b+c/x}-1}}\right]\left(e^{-ax\sqrt{2^{b+c/x}-1}}\right)=0$$ where $a,b,c,d$ are all positive constants. I am unable to solve it. Please help me in getting there. Many thanks in advance.

Answer 1

First notice that the function under limit is not defined in the left neighborhood of $0$, because assuming $x<0$, the expression under square root has to be non negative, i.e. $2^{b+c/x}-1 \geq 0$, and you can put these together to show that $x \leq -c/b$ then. So I'll assume you want to compute limit for $x \to 0^+$.

Next thing, and it is not necessary but personally I find it more intuitive, is to replace $x$ with $\frac{1}{x}$ and calculate the limit as $x\to\infty$.

After this and several algebraic manipulations, you would get

$$ \lim_{x \to \infty} -cd\frac{\ln 2}{2} f(x)g(x) $$

where $$f(x) = \frac{a}{x}-\frac{2a}{c\ln 2 x^2}+\frac{a}{x(2^{b+cx}-1)}+\frac{1}{\sqrt{2^{b+cx}-1}}+\frac{1}{(2^{b+cx}-1)^{3/2}}$$ $$g(x) = \frac{x^2}{\mathrm{e}^{a\frac{\sqrt{2^{b+cx}}}{x}}}$$

Now it is enough to show that $\lim_{x \to \infty} f(x) = 0$ and $\lim_{x \to \infty} g(x) = 0$.

The $\lim_{x \to \infty} f(x) = 0$ is straightforward since each of the fractions converge to $0$ (notice that $2^{b+cx}-1 \to \infty$ as $x \to \infty$).

The $\lim_{x \to \infty} g(x) = 0$ is a bit more tricky, but notice that $\sqrt{2^{b+cx}}$ grows faster than $x^2$, i.e. for sufficiently large $x$ you have

$$ \sqrt{2^{b+cx}} \geq x^2 $$

Applying it to the $g(x)$, you get

$$ 0 \leq \frac{x^2}{\mathrm{e}^{a\frac{\sqrt{2^{b+cx}}}{x}}} \leq \frac{x^2}{\mathrm{e}^{ax}} $$

Now it is clear the right side converge to $0$ (since exponential function grows faster than any polynomial, same principle as we just used), therefore the left side also converge to $0$.

Answer 2

Try something simpler. Get rid of most of the constants. Instead of $x\to 0^+,$ replace $x$ by $1/x$ and let $x\to \infty.$ (For me it's easier to think this way.) Throw away the $1$'s you keep subtracting, they're nothing compared with $2^x.$ So here's what I looked at:

$$\tag 1 \frac{1}{2^x}\left [ 2^x + x2^x + x^22^{x/2}\right ]e^{-2^{x/2}/x}.$$

That's a lot less than

$$3x^22^xe^{-2^{x/2}/x}.$$

Now apply $\ln$ to get

$$\ln 3 + 2 \ln x + x\ln 2 - 2^{x/2}/x.$$

That has to go to $-\infty$ because of the exponential growth of $2^{x/2}.$ That tells me that $(1)\to 0.$ And that's pretty good evidence that your original expression $\to 0.$ Now you have an idea where you're going with that messy thing.